# Alexandrov spaces which are not limits of Riemannian manifolds

Are there important/ interesting/ natural examples of compact Alexandrov spaces with curvature bounded from below which are not Gromov-Hausdorff limits of smooth compact Riemannian manifolds with uniformly bounded from below sectional curvature? (The condition of compactness might be relaxed somehow.)

EDIT: The dimension of manifolds in the approximating sequence is supposed to be bounded from above.

• Can you get a tripod in such a way? (tripod is the 1-skeleton of the 4-vertices graph given by 1 vertex liked to the three others) – YCor Jun 3 '15 at 8:25
• @YCor: Tripod has curvature bounded from above but not from below. – MKO Jun 3 '15 at 8:33
• Ah OK (I saw "bounded from below" only in the second part of the sentence). – YCor Jun 3 '15 at 8:52
• What about the metric suspension over $\mathbb{RP}^2$. – foliations Jun 3 '15 at 14:44

• There is an Alexandrov space, say $A$, with curvature $\ge 1$ which can not obtained as a limit of Riemannian manifolds with curvature $\ge \kappa$, if $\kappa>\tfrac14$; see "Metric constraints on exotic spheres via Alexandrov geometry." by Grove and Wilhelm.

• There is an Alexandrov space, say $A$, such that if it can appear as a limit of Riemannian manifolds $M_n$ with uniformly bounded curvature then $\dim M_n\ge \dim A+8$ for all large $n$; see "Regularity of limits of noncollapsing sequences of manifolds" by Kapovitch. It is expected that in this formula one can exchange $8$ to $\infty$.

• Is the space $A$ compact in each example? In the second example, the curvature of $M_n$ is uniformly bounded from below or from both sides? – MKO Jun 3 '15 at 13:56
• (1) yes always, (2) the construction is local and one can assume $A$ is compact, – Anton Petrunin Jun 4 '15 at 10:10
• Sorry, I did not understand your answer. What is "yes always"? Is it related to the fact that the bounds on curvature are always lower, or that $A$ is always compact? Could you please elaborate. Thanks. – MKO Jun 4 '15 at 11:09
• in (2) you get examples which are compact and which are not compact. – Anton Petrunin Jun 5 '15 at 13:54

I believe (I am not an expert in the subject) a simple example (though using a highly non-trivial theorem) is given by considering the following example: Let $$\mathbb{S}^3\subset \mathbb{R}^4$$ be the round sphere of curvature $1$ and consider the isometry $\phi:\mathbb{S}^3\to \mathbb{S}^3$ induced by $$(x_1, x_2, x_3, x_4)\mapsto (x_1, -x_2, -x_3, -x_4)$$ and let $$X=\mathbb{S}^3/\phi.$$ be the obvious quotient space with the natural quotient metric. This quotient has curvature $\geq 1$, but is not a topological manifold (as small balls about $(\pm 1 , 0, 0, 0)$ look like cones over $\mathbb{RP}^2$).

Hence, by Perelman's stability theorem, $X$ cannot be the GH limit of smooth manifolds of curvature $\geq k$ for any $k\leq 1$.

• The stability theorem concerns the situation when there is no collapse. In the situation of my question collapse is possible. – MKO Jun 3 '15 at 15:14